derivative f(x)=(x^2+1)(2-x)

解答

导数

f (x) = (x^{2} + 1) (2 - x)

4 x - 3 x^{2} - 1

求解步骤

\frac{d}{d x} ((x^{2} + 1) (2 - x))

使用微分乘法定则:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{2} + 1, g = 2 - x

= \frac{d}{d x} (x^{2} + 1) (2 - x) + \frac{d}{d x} (2 - x) (x^{2} + 1)

\frac{d}{d x} (x^{2} + 1) = 2 x

\frac{d}{d x} (2 - x) = - 1

= 2 x (2 - x) + (- 1) (x^{2} + 1)

化简

2 x (2 - x) + (- 1) (x^{2} + 1) : 4 x - 3 x^{2} - 1

= 4 x - 3 x^{2} - 1

x \to - 3 lim (2 x + 5)

\frac{d}{d x} (cos (x^{- 6}))

t an g e n t f (x) = 8 e^{x} + x, at x = 0

\int (x + 1) e^{3 x + 1} d x

\int 5 x^{4} - 2 x^{5} d x